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Shear Force and Bending Moment - Solved example for overhanging beam

Problem 5-3

Calculate the values and draw the diagrams for shearing force and bending moment for the following overhanging beam shown in figure 5-3(a).

Figure 5-3(a)

Solution:

The free-body-diagram is shown in figure 5-3(b). The unknown reaction components of the beam are calculated in Prob 4-2 as  Ax=0,  Ay = -10 kN and By = 55 kN;

Shearing force calculations

In case of point load acting at a point, we should calculate shear force on both sides (left and right) of the point.

FA-left = 0;

FA-right = -10 kN;

FC = -10 kN;

FB-left = -10 – 5x4 = - 30 kN;

FB -right = -10 – 5x4 + 55 = +25 kN;

FD-left = +25 kN;

FD-right = +25-25 = 0;

The values of shear force are plotted in SFD in figure 5-3(b)

Bending moment calculations

MA = 0;

MC = -10x2 = -20 kNm;

MB = -10x6 – 5x4x2 = -60 -40 = -100 kNm;

MD = 0;

The values of bending moment are plotted in BMD.in figure 5-3(b)

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Figure 5-3(b)

The bending moment diagram shows that bending moment is hogging throughout the span of the beam and maximum BM is at B (shear force changes sign at B). The bending moment diagram is straight line from A to C (due to point load), parabolic from C to B (due to udl) and straight line from B to D (due to point load).  It is important to note that the parabola is a increasing parabola (means slope is increasing) because the shear force is increasing from C to B. Rate of change of bending moment is equal to shear force.

More examples on shear and moment diagrams

Example 5-1  Example 5-2   Example 5-4

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